Question

Find the equation of plane perpendicular to the yz plane and passing through points (2,3,1) and (4,-5,3)

Answer 1

let the plane be ... ax+by+cz+d=0

put the values of x and y,
2a+3b+c+d=0
4a-5b+3c+d=0

and for perpendicular planes, normals are perpendicular
<a,b,c><0,0,1>=0
c=0

Answer 2

oh i dont think i is n.v ,sry...let me see..

Answer 3

i think, more simpler ....
2a+3b+c=1
4a-5b+3c=1

Answer 4

passing through points (2,3,1) and (4,-5,3) means the vector v(2,-8,2) is paralel to the plane.Equation of yz plane is : x=0. So perpendicular vector to it is:p(1,0,0)
B=vXp this is the vector perpendicular to the plane you looking for (x represents vector product)

Answer 5

and normal of yz plane



I think it should be like this